Find the equation of the tangent line to the curve (a lemniscate) 2(x^2+y^2)2=25(x^2−y^2) at the point (−3,1). The equation of this tangent line can be written in the form y=mx+b where m is:

- anonymous

- katieb

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

First Differentiate the function to find the slope.

- anonymous

thank you but i dont know process. i am not sure about calculation, particulary differentiation

- anonymous

I mean i know the process

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- zepdrix

You know the process, but you need assistance with the differentiation portion?

- anonymous

Yes, when it comes to both chain rule and products. I get really confused

- zepdrix

\[\large 2(x^2+y^2)^2=25(x^2−y^2)\]So we'll be differentiating this `implicitly`, meaning, `as is`. No solving for y ahead of time.
When we execute the chain rule, a few y' will pop out here and there, and we'll want to solve for that variable.
\(\large y'(-3,1)=m\). This is the m in our equation \(\large y=mx+b\).
Hmm so let's see...

- zepdrix

It appears we have no product rule which is nice.

- anonymous

OK, Y7=(-3,1)=m. right?

- zepdrix

We should be able to get a numerical value for m, no x's or y's in our m.
I'm not really sure what that Y7 is.

- anonymous

Ops my mistake

- zepdrix

Taking the derivative of the left side gives us, `taking the derivative of the outermost function first, the square on the brackets`:\[\large 2\cdot2(x^2+y^2)^1\color{royalblue}{\frac{d}{dx}(x^2+y^2)}\]The blue is due to the chain rule.
Understand that part?

- anonymous

yes, so far good.

- zepdrix

\[\large 2\cdot2(x^2+y^2)\color{royalblue}{(2x+2yy')}\]So we get something like this on the left side.

- anonymous

i lied, I know product rule, power rule etc, but i dont know what the chain rule is

- zepdrix

lol :)

- zepdrix

mmmm i always have trouble explaining that one. It's very unintuitive.

- zepdrix

hmm

- anonymous

could you give me example of chain rule problem maybe?

- zepdrix

The definition says this,\[\large \frac{d}{dx}f\left(g(x)\right) \qquad = \qquad f'(g(x))\cdot g'(x)\]When you take the derivative of this, a function within a function, you
~Differentiate the outer function, f in this case.
~Then you make a copy of the inner function, and multiply by it's derivative.
Here is a silly example.
\[\large \frac{d}{dx}\left(x^2\right)^{4}\]
In this example we're thinking of \(\large g(x)=x^2\).
And \(\large f(x)=x^4\).
If you remember back to doing compositions of functions, you would understand that when we plug g(x) INTO f(x), our f(x), \(\large f(\color{royalblue}{x})=\left(\color{royalblue}{x}\right)^4\) will become, \(\large f(\color{royalblue}{g(x)})=\left(\color{royalblue}{g(x)}\right)^4\) which is \(\large f(\color{royalblue}{x^2})=\left(\color{royalblue}{x^2}\right)^4\).
Applying the chain rule gives us,\[\large f'(g(x))=4\left(g(x)\right)^3 g'(x)\]See how the rule tells us to make a copy of the inner function and take it's derivative?\[\large f'(g(x))=4\left(x^2\right)^3 (x^2)'\]\[\large f'(g(x))=4\left(x^2\right)^3 (2x)\]

- anonymous

Oo.. I see, Thank you. It was very clear! really cool :)

- zepdrix

If you were to simplify this down, you would get,\[\large f'(g(x))=8x^7\]
Which we know is correct, because we could have easily approached this example without using the chain rule.\[\large y=(x^2)^4=x^8\]And we know the derivative will match what we got from the chain rule example.

- zepdrix

It's a weird concept :d you're like.. making copies of stuff.. but teachers never use that type of language to describe it. So I dunno lol

- anonymous

no, you explain well. i always get confused and could not tell wether i should use chain or product.

- zepdrix

So we applied the derivative of the left side,\[\large \frac{d}{dx}(thing)^2 \qquad \rightarrow \qquad 2(thing)(thing)'\]
right? :O

- anonymous

haha thats even more clear haha def, taking note

- zepdrix

lol :d

- anonymous

could we back to the first question?

- anonymous

i took note

- zepdrix

This one? :o
\[\large 2(x^2+y^2)^2=25(x^2−y^2)\]

- anonymous

yes, haha

- anonymous

I see the chain rule now,
(x^2+y^2)^2?

- zepdrix

\[\large 2(x^2+y^2)\]That's the derivative of the outer function yes? The square on the brackets. But the chain rule is telling us to make a copy of the inside, and take it's derivative.\[\large 2(x^2+y^2)(x^2+y^2)'\]

- zepdrix

\[\large 2(x^2+y^2)(2x+2y)\]So we should get somethingggggggg like this, yes?

- anonymous

:O why two in the fron of the bracket? I understand the (2x+2y) tho

- zepdrix

You applied the power rule to the outer function \(\large \left(\qquad \right)^2\)

- anonymous

power rule and chain double? :O

- zepdrix

Power rule the outside, make a copy of the inside, take derivative of copy.

- anonymous

Ok, i understand

- zepdrix

Since we took our derivative with respect to x, any time we differentiate a variable that IS NOT x, like y for example, we have to apply the chain rule, and multiply by y'.

- zepdrix

\[\large 2(x^2+y^2)(2x+2y\color{royalblue}{y'})\]See this blue term?
We took the derivative of a variable that is not x, namely y, so a y' popped out.

- anonymous

mmmmmm.... Ok. I think :/

- anonymous

so whenever its chain rule i put y' in the end?

- zepdrix

No. Let's not think of that portion as chain rule, I think that's just going to confuse you.
Whenever we take the derivative of y, we will multiply by y'.
We don't throw it on the end, we always attach it to the y that we're taking the derivative of.

- anonymous

right, Ok. i will remember that by my heart.

- zepdrix

lol c:

- anonymous

haha

- anonymous

whats nxt stp?

- zepdrix

Try taking the derivative of the right side, tell me what you get.

- anonymous

Ok!

- zepdrix

Ignore the 25 in front. No chain rule, just take the derivative of the stuff in the brackets.

- anonymous

25(2x-2y)?

- zepdrix

close, but you took the derivative of a y, you need to multiply that part by y'

- anonymous

25(2x-2yy')?

- anonymous

its still funny how i have to put y' in the end. is it for implicit derviation?

- zepdrix

yes.
If we had an equation like this,\[\large y=x^2\]The derivative gives us,\[\large y'=2x\]Ok that makes sense.. The y' is telling us what the derivative of y is. But what if we had something like this?\[\large y^2=x^2\]Taking the derivative,\[\large 2y=2x\] Hmm we don't have a derivative term do we?

- anonymous

Ok, thats cool

- anonymous

so 2(x^2+y~2)(2x+2y)=25(2x-2yy')

- zepdrix

Wasn't there a 2 in front of the equation on the left at the start?? Or was that just a typo?

- zepdrix

2(x^2+y^2)^2=25(x^2−y^2)

- zepdrix

See that 2 at the front? That would give us an extra factor of 2 in our derivative.

- anonymous

the question equation is
2(x^2+y^2)^2=25(x^2-y^2)

- anonymous

so is it 2*2?

- zepdrix

yes

- anonymous

\[4(x^2+y^2)(2x+2y)=25(2x-2yy')\]

- zepdrix

\[\large 4(x^2+y^2)(2x+2y\color{royalblue}{y'})=25(2x-2yy')\]Woops you missed a y'.

- anonymous

ops! haha thats derivative of the first equation?

- zepdrix

yes.
We took the derivative of y on each side, so each y will get a y'

- anonymous

Ok,

- zepdrix

Hmmm from here, I think it's going to be easier to FIRST plug in your coordinate \(\large (-3,1\;)\) and THEN solve for y'.

- anonymous

plug both number into the equation?

- zepdrix

yes, -3 for all the x's. 1 in for all the y's.

- zepdrix

and don't plug anything in for y'. That's what we'll be solving for.

Looking for something else?

Not the answer you are looking for? Search for more explanations.